02.3.1 Understanding Function Parity

A function describes a relationship between an input and an output.

For example:

If we substitute \(x=3\):

If we substitute \(x=-3\):

Notice that both inputs produce the same output:

This is not a coincidence. It happens because squaring a positive or negative number produces the same nonnegative result.

For every real number \(x\):

Therefore:

This function is even.

Now consider another function:

Substituting \(x=2\):

Substituting \(x=-2\):

Notice that:

This relationship holds for every real number:

Therefore:

This function is odd.

What does parity tell us?

Parity describes the relationship between a function’s outputs at opposite inputs.

Instead of studying every positive and negative input separately, we can use symmetry to understand the function’s entire behavior.

However, a function must be defined at both \(x\) and \(-x\) for the comparison to apply throughout its domain.

02.3.2 Even Functions

A function \(f\), with domain \(D\), is called even if:

1. Whenever \(x\in D\), then \(-x\in D\).

2. For every \(x\in D\):

   
   f (−x) = f (x)
   

   The essential formula is \(f(-x)=f(x)\). The empty box is only a formatting illustration; the Tutor LMS-compatible boxed formula appears below.

   In simple terms, changing the sign of the input does not change the output.

   Example 1: The function \(f(x)=x^2\)

   Determine whether:

   \[ f(x)=x^2 \]

   is even.

   Step 1. Replace \(x\) with \(-x\).

   \[ f(-x)=(-x)^2 \]

   Step 2. Simplify.

   \[ f(-x)=x^2 \]

   Step 3. Compare with the original function.

   Since:

   \[ f(x)=x^2 \]

   we obtain:

   \[ f(-x)=f(x) \]

   Conclusion: The function is even.

   Example 2: An even polynomial

   Determine the parity of:

   \[ f(x)=3x^4-2x^2+7 \]

   Replace \(x\) with \(-x\):

   \[ f(-x)=3(-x)^4-2(-x)^2+7 \]

   Since:

   \[ (-x)^4=x^4 \]

   and:

   \[ (-x)^2=x^2 \]

   we obtain:

   \[ f(-x)=3x^4-2x^2+7 \]

   This is identical to the original function.

   Therefore:

   \[ f(-x)=f(x) \]

   Final answer: Even function.

   Key properties of even functions

   For an even function:

   • Opposite inputs produce equal outputs.

   • The graph is symmetric about the y-axis.

   • Replacing \(x\) with \(-x\) leaves the function unchanged.

   • A nonzero constant function is even when its domain is symmetric about zero.

   Common examples include:

   \[ f(x)=x^2 \]

   \[ f(x)=x^4 \]

   \[ f(x)=|x| \]

   \[ f(x)=\cos x \]

   All four functions are even on their standard real domains.

    

   02.3.3 Odd Functions

   A function \(f\), with domain \(D\), is called odd if:

   1. Whenever \(x\in D\), then \(-x\in D\).

   2. For every \(x\in D\):

   \[ f(-x)=-f(x) \]

   In simple terms, changing the sign of the input changes the sign of the output.

   For example, if:

   \[ f(4)=10 \]

   and \(f\) is odd, then:

   \[ f(-4)=-10 \]

   Example 3: The function \(f(x)=x^3\)

   Determine whether:

   \[ f(x)=x^3 \]

   is odd.

   Step 1. Replace \(x\) with \(-x\).

   \[ f(-x)=(-x)^3 \]

   Step 2. Simplify.

   \[ f(-x)=-x^3 \]

   Step 3. Compare with the negative of the original function.

   The original function is:

   \[ f(x)=x^3 \]

   Its negative is:

   \[ -f(x)=-x^3 \]

   Therefore:

   \[ f(-x)=-f(x) \]

   Final answer: Odd function.

   Example 4: An odd polynomial

   Determine the parity of:

   \[ f(x)=2x^5-3x^3+x \]

   Replace \(x\) with \(-x\).

   \[ f(-x)=2(-x)^5-3(-x)^3+(-x) \]

   Simplify:

   \[ f(-x)=-2x^5+3x^3-x \]

   Now calculate:

   \[ -f(x)=-(2x^5-3x^3+x) \]

   Distribute the negative sign:

   \[ -f(x)=-2x^5+3x^3-x \]

   Therefore:

   \[ f(-x)=-f(x) \]

   Conclusion: The function is odd.

   Important property of odd functions

   If an odd function is defined at \(x=0\), then:

   \[ f(0)=0 \]

   Why?

   For an odd function:

   \[ f(-x)=-f(x) \]

   Substitute \(x=0\):

   \[ f(0)=-f(0) \]

   Add \(f(0)\) to both sides:

   \[ 2f(0)=0 \]

   Therefore:

   \[ f(0)=0 \]

   This means that the graph of an odd function passes through the origin if zero belongs to its domain.

   However, not every odd function is defined at zero.

   For example:

   \[ f(x)=\frac1x \]

   is odd on its domain \(\mathbb R\setminus\{0\}\), but is undefined at \(x=0\).

   Common odd functions

   \[ f(x)=x \]

   \[ f(x)=x^3 \]

   \[ f(x)=\frac1x \]

   \[ f(x)=\sin x \]

   The last two examples illustrate that an odd function does not need to be a polynomial.

    

   02.3.4 Graphical Symmetry of Even and Odd Functions

   The algebraic definitions of even and odd functions correspond to specific graphical properties.

   Even functions — Symmetry about the y-axis

   For an even function:

   \[ f(-x)=f(x) \]

   If the point:

   \[ (a,b) \]

   belongs to the graph, then:

   \[ (-a,b) \]

   also belongs to the graph.

   For example, the graph of:

   \[ y=x^2 \]

   contains:

   \[ (2,4) \]

   It must also contain:

   \[ (-2,4) \]

   These points are reflections of one another across the y-axis.

   Odd functions — Symmetry about the origin

   For an odd function:

   \[ f(-x)=-f(x) \]

   If:

   \[ (a,b) \]

   belongs to the graph, then:

   \[ (-a,-b) \]

   also belongs to the graph.

   For example, the graph of:

   \[ y=x^3 \]

   contains:

   \[ (2,8) \]

   It must also contain:

   \[ (-2,-8) \]

   These two points are related by a 180-degree rotation around the origin.

   Visual comparison

   Graphical Symmetry of Functions

    

   Function type	Symmetry	Paired points
   Even	Y-axis reflection	(a, b) and (−a, b)
   Odd	180° rotation about origin	(a, b) and (−a, −b)
   Neither	Neither symmetry required	No general rule

   Understanding Even and Odd Symmetry

   Even Function

   f(x) = x²

   (+2) → 4

   (−2) → 4

   Same output

   Symmetric about the y-axis

   Odd Function

   g(x) = x³

   (+2) → +8

   (−2) → −8

   Opposite outputs

   Symmetric about the origin

   02.3.5 Functions That Are Neither Even Nor Odd

   Not every function has even or odd symmetry.

   A function is classified as neither even nor odd if it does not satisfy either parity condition throughout its domain.

   Consider:

   \[ f(x)=x^2+x \]

   Example 5: A function that is neither even nor odd

   Step 1. Replace \(x\) with \(-x\).

   \[ f(-x)=(-x)^2+(-x) \]

   Simplify:

   \[ f(-x)=x^2-x \]

   Step 2. Compare with the original function.

   Original:

   \[ f(x)=x^2+x \]

   Since:

   \[ x^2-x\neq x^2+x \]

   as identities in \(x\), the function is not even.

   Step 3. Compare with the negative of the original function.

   \[ -f(x)=-(x^2+x) \]

   \[ -f(x)=-x^2-x \]

   But:

   \[ f(-x)=x^2-x \]

   Therefore:

   \[ f(-x)\neq-f(x) \]

   as identities.

   Final answer: Neither even nor odd.

   Why does this happen?

   The expression contains both:

   • \(x^2\), which is an even function.

   • \(x\), which is an odd function.

   Their sum does not have either symmetry.

   For example:

   \[ f(2)=2^2+2=6 \]

   \[ f(-2)=(-2)^2-2=2 \]

   These outputs are neither equal nor opposites.

   Example 6: Another function with neither parity

   Consider:

   \[ f(x)=x^3+x^2 \]

   Substitute \(-x\):

   \[ f(-x)=(-x)^3+(-x)^2 \]

   \[ f(-x)=-x^3+x^2 \]

   The original function is:

   \[ f(x)=x^3+x^2 \]

   Its negative is:

   \[ -f(x)=-x^3-x^2 \]

   We can see that:

   \[ f(-x)\neq f(x) \]

   and:

   \[ f(-x)\neq-f(x) \]

   Therefore, the function is neither even nor odd.

   Important: Identifying both even-power and odd-power terms is a useful shortcut for polynomials, but the complete algebraic test remains the reliable method for all functions.

    

   02.3.6 The Importance of a Symmetric Domain

   One of the most frequently overlooked conditions in parity problems is the domain of the function.

   An even or odd function must have a domain that is symmetric about zero.

   What is a symmetric domain?

   A domain \(D\) is symmetric about zero if:

    

   x ∈ D
   ⟹
   −x ∈ D

   For example:

   \[ D=[-4,4] \]

   is symmetric about zero.

   If \(x=3\) belongs to the domain, then \(x=-3\) also belongs to the domain.

   Similarly:

   \[ D=(-\infty,\infty) \]

   is symmetric about zero.

   And:

   \[ D=(-\infty,-2)\cup(2,\infty) \]

   is also symmetric about zero.

   However:

   \[ D=[0,\infty) \]

   is not symmetric about zero.

   For example, 3 belongs to this domain, but −3 does not.

   Example 7: A familiar formula with a restricted domain

   Consider:

   \[ f(x)=x^2,\qquad x\in[0,4] \]

   Normally, \(f(x)=x^2\) is an even function on \(\mathbb R\).

   But here the domain is restricted to:

   \[ [0,4] \]

   The value \(x=2\) belongs to the domain.

   However, \(x=-2\) does not.

   Therefore, \(f(-2)\) is not defined for this restricted function.

   The function cannot satisfy the even or odd parity definition over its entire domain.

   Final answer: Neither even nor odd on the specified domain.

   This is different from the function:

   \[ g(x)=x^2,\qquad x\in[-4,4] \]

   which is even.

   Example 8: A rational function with a symmetric domain

   Consider:

   \[ f(x)=\frac1x \]

   Its domain is:

   \[ D=\mathbb R\setminus\{0\} \]

   This domain is symmetric about zero.

   Now substitute \(-x\):

   \[ f(-x)=\frac1{-x} \]

   \[ f(-x)=-\frac1x \]

   Therefore:

   \[ f(-x)=-f(x) \]

   Final answer: Odd function.

   Notice that 0 does not need to belong to the domain for a function to be odd.

   Symmetric vs. non-symmetric domains

   Symmetric and Non-Symmetric Domains

   Domain	Symmetric about zero?
   ℝ	Yes
   [−5, 5]	Yes
   (−3, 3)	Yes
   (−∞, −1) ∪ (1, ∞)	Yes
   [0, ∞)	No
   [−2, 5]	No
   ℝ ∖ {2}	No

   Essential principle: A symmetric domain is necessary for a function to be even or odd, but a symmetric domain alone does not prove either parity.

    

   02.3.7 Algebraic Method for Testing Function Parity

   The most reliable method for determining parity is an algebraic substitution.

   General procedure

   How to Determine Function Parity

   Step 1: Check the domain

   The domain must be symmetric about zero.

   ↓

   Step 2: Substitute -x

   Calculate f(-x) and simplify.

   ↓

   Step 3: Test for even parity

   f(−x) = f(x)

   ↓

   Step 4: Test for odd parity

   f(−x) = −f(x)

   ↓

   Step 5: State the classification

   Even, odd, both, or neither.

   Example 9: Determine parity using the full procedure

   Consider:

   \[ f(x)=x^4-5x^2+4 \]

   Step 1. The domain is all real numbers, which is symmetric about zero.

   Step 2. Calculate \(f(-x)\).

   \[ f(-x)=(-x)^4-5(-x)^2+4 \]

   Step 3. Simplify.

   \[ f(-x)=x^4-5x^2+4 \]

   Step 4. Compare with the original function.

   \[ f(x)=x^4-5x^2+4 \]

   Therefore:

   \[ f(-x)=f(x) \]

   Final answer: Even function.

   Example 10: An odd rational function

   Determine the parity of:

   \[ f(x)=\frac{x}{x^2+1} \]

   Step 1. Determine the domain.

   Since:

   \[ x^2+1>0 \]

   for every real \(x\), the domain is:

   \[ D=\mathbb R \]

   Step 2. Calculate \(f(-x)\).

   \[ f(-x)=\frac{-x}{(-x)^2+1} \]

   Simplify:

   \[ f(-x)=\frac{-x}{x^2+1} \]

   Factor out −1:

   \[ f(-x)=-\frac{x}{x^2+1} \]

   Therefore:

   \[ f(-x)=-f(x) \]

   Final answer: Odd function.

   Example 11: A rational function that is neither

   Consider:

   \[ f(x)=\frac1{x-2} \]

   The domain is:

   \[ D=\mathbb R\setminus\{2\} \]

   This domain is not symmetric about zero.

   For example, \(-2\) is in the domain but \(2\) is not.

   Therefore, the function is neither even nor odd on its natural domain.

   There is no need to continue with algebraic parity testing.

    

   02.3.8 Common Polynomial and Elementary Function Examples

   It is useful to recognize the parity of common functions quickly, especially in multiple-choice questions.

   However, these classifications assume each function is used on its natural real domain unless otherwise stated.

   Polynomial examples

   Common Function Parity Examples

   Function	Parity	Reason
   f(x) = x2	Even	Even power
   f(x) = x3	Odd	Odd power
   f(x) = x4 + 2x2 + 1	Even	Even powers and constant
   f(x) = x5 − 3x	Odd	Odd powers
   f(x) = x2 + x	Neither	Mixed powers
   f(x) = 5	Even	Nonzero constant
   f(x) = 0	Both	Both identities hold

   Classifications assume the natural real domain of each function.

   Important rule for polynomial functions

   A polynomial containing only even powers of \(x\), including a constant term, is even.

   For example:

   \[ f(x)=4x^6+3x^4-2x^2+8 \]

   A polynomial containing only odd powers of \(x\), with no nonzero constant term, is odd.

   For example:

   \[ g(x)=2x^7-5x^3+x \]

   A polynomial containing nonzero terms of both even and odd powers is neither even nor odd.

   For example:

   \[ h(x)=x^4+2x^3-x^2 \]

   These rules follow because polynomial functions have the entire real number line as their domain.

    

   02.3.9 Parity of Basic Trigonometric Functions

   Trigonometric functions also have parity properties.

   Three important examples are sine, cosine, and tangent.

   These functions will be studied in more detail in Lesson 02.7. Here, we focus only on their symmetry.

   Sine is an odd function

   The sine function satisfies:

   \[ \sin(-x)=-\sin x \]

   Therefore:

   \[ f(x)=\sin x \]

   is odd.

   Its graph is symmetric about the origin.

   Cosine is an even function

   The cosine function satisfies:

   \[ \cos(-x)=\cos x \]

   Therefore:

   \[ f(x)=\cos x \]

   is even.

   Its graph is symmetric about the y-axis.

   Tangent is an odd function

   The tangent function satisfies:

   \[ \tan(-x)=-\tan x \]

   whenever tangent is defined.

   Therefore:

   \[ f(x)=\tan x \]

   is odd on its natural domain:

   \[ x\neq\frac{\pi}{2}+k\pi,\qquad k\in\mathbb Z \]

   The excluded values occur in symmetric pairs, so the natural domain is symmetric about zero.

   Trigonometric parity summary

   Trigonometric Function Parity

   Function	Parity	Identity
   sin x	Odd	sin(−x) = −sin(x)
   cos x	Even	cos(−x) = cos(x)
   tan x	Odd	tan(−x) = −tan(x)

   Example 12: A combination of sine and cosine

   Determine the parity of:

   \[ f(x)=\sin x+\cos x \]

   Substitute \(-x\):

   \[ f(-x)=\sin(-x)+\cos(-x) \]

   Apply the trigonometric identities:

   \[ f(-x)=-\sin x+\cos x \]

   The original function is:

   \[ f(x)=\sin x+\cos x \]

   Its negative is:

   \[ -f(x)=-\sin x-\cos x \]

   The substituted expression is neither identical to \(f(x)\) nor to \(-f(x)\) for all real \(x\).

   Final answer: Neither even nor odd.

    

   02.3.10 Special Case: A Function Can Be Both Even and Odd

   Students sometimes assume that every function must belong to exactly one of three categories: even, odd, or neither.

   However, there is a fourth possibility.

   A function can be both even and odd.

   Consider:

   \[ f(x)=0 \]

   for every real \(x\).

   First, test the even condition:

   \[ f(-x)=0=f(x) \]

   The function is even.

   Now test the odd condition:

   \[ f(-x)=0=-f(x) \]

   The function is also odd.

   Therefore, the zero function is both even and odd.

   Why is the zero function special?

   Suppose a function is both even and odd on a symmetric domain.

   Then:

   \[ f(-x)=f(x) \]

   and:

   \[ f(-x)=-f(x) \]

   Combining these two relationships gives:

   \[ f(x)=-f(x) \]

   Therefore:

   \[ 2f(x)=0 \]

   Hence:

   \[ f(x)=0 \]

   for every input in its domain.

   Thus, the only real-valued function that is simultaneously even and odd is the zero function on its domain.

   Comparison of classifications

   Function	Classification
   f(x) = x2	Even only
   f(x) = x3	Odd only
   f(x) = x2 + x	Neither
   f(x) = 0	Both

   02.3.11 Advanced Worked Examples — Applying Parity

   These examples combine algebraic manipulation, domain analysis, and parity properties.

   Example 13: Even function with a square root

   Determine the parity of:

   \[ f(x)=\sqrt{9-x^2} \]

   Step 1. Determine the domain.

   The expression inside the square root must be nonnegative.

   \[ 9-x^2\geq0 \]

   Therefore:

   \[ x^2\leq9 \]

   \[ -3\leq x\leq3 \]

   The domain is:

   \[ [-3,3] \]

   This domain is symmetric about zero.

   Step 2. Calculate \(f(-x)\).

   \[ f(-x)=\sqrt{9-(-x)^2} \]

   Simplify:

   \[ f(-x)=\sqrt{9-x^2} \]

   Therefore:

   \[ f(-x)=f(x) \]

   Final answer: Even function.

   Example 14: A function involving absolute value

   Determine the parity of:

   \[ f(x)=x|x| \]

   Recall that \(|x|\) represents the absolute value of \(x\).

   For example:

   \[ |-3|=3 \]

   And:

   \[ |3|=3 \]

   Thus:

   \[ |-x|=|x| \]

   Now calculate:

   \[ f(-x)=(-x)|-x| \]

   Using \(|-x|=|x|\):

   \[ f(-x)=-x|x| \]

   Therefore:

   \[ f(-x)=-f(x) \]

   Final answer: Odd function.

   Example 15: Finding an unknown coefficient

   Find the value of \(k\) that makes the function:

   \[ f(x)=x^3+kx^2-4x \]

   odd.

   Step 1. Calculate \(f(-x)\).

   \[ f(-x)=(-x)^3+k(-x)^2-4(-x) \]

   Simplify:

   \[ f(-x)=-x^3+kx^2+4x \]

   Step 2. Calculate \(-f(x)\).

   \[ -f(x)=-x^3-kx^2+4x \]

   Step 3. Apply the odd-function condition.

   For the function to be odd:

   \[ f(-x)=-f(x) \]

   Therefore:

   \[ -x^3+kx^2+4x=-x^3-kx^2+4x \]

   Subtract the common terms:

   \[ kx^2=-kx^2 \]

   \[ 2kx^2=0 \]

   For this to hold for every real \(x\), we require:

   \[ k=0 \]

   Final answer:

   \[ k=0 \]

   When \(k=0\), the function becomes:

   \[ f(x)=x^3-4x \]

   which is odd.

   Example 16: Using parity to find an unknown output

   An even function satisfies:

   \[ f(5)=-8 \]

   Find \(f(-5)\).

   Because the function is even:

   \[ f(-x)=f(x) \]

   Therefore:

   \[ f(-5)=f(5) \]

   Final answer:

   \[ f(-5)=-8 \]

   Now consider an odd function \(g\) satisfying:

   \[ g(5)=-8 \]

   For an odd function:

   \[ g(-x)=-g(x) \]

   Therefore:

   \[ g(-5)=-g(5) \]

   \[ g(-5)=-(-8)=8 \]

   Final answer:

   \[ g(-5)=8 \]

   These questions can be solved without knowing the complete formula of the function.

    

   02.3.12 Common Mistakes to Avoid

   Mistake 1: Confusing \(f(-x)\) with \(-f(x)\)

   These expressions have different meanings.

   Consider:

   \[ f(x)=x^2+3 \]

   Then:

   \[ f(-x)=(-x)^2+3=x^2+3 \]

   But:

   \[ -f(x)=-(x^2+3)=-x^2-3 \]

   Replacing the input with its negative is not the same as multiplying the output by −1.

   Mistake 2: Assuming every function containing \(x^2\) is even

   Consider:

   \[ f(x)=x^2+x \]

   The presence of \(x^2\) does not make the entire function even.

   The additional \(x\) term destroys even symmetry.

   Mistake 3: Ignoring the domain

   Consider:

   \[ f(x)=\sqrt{x} \]

   Its natural real domain is:

   \[ [0,\infty) \]

   This domain is not symmetric about zero.

   Therefore, the function is neither even nor odd.

   Mistake 4: Thinking every odd function passes through the origin

   An odd function satisfies \(f(0)=0\) only if it is defined at zero.

   For example:

   \[ f(x)=\frac1x \]

   is odd, but undefined at zero.

   Mistake 5: Assuming \(f(0)=0\) proves that a function is odd

   Consider:

   \[ f(x)=x^2 \]

   It satisfies:

   \[ f(0)=0 \]

   But it is even, not odd.

   Therefore, \(f(0)=0\) alone is not sufficient to establish odd parity.

   Mistake 6: Testing only one pair of inputs

   Suppose a function happens to satisfy:

   \[ f(-2)=f(2) \]

   This alone does not prove that the function is even.

   To establish even parity, the equality must hold for every input in a symmetric domain.

   A single counterexample, however, is enough to disprove a proposed parity identity.

   Mistake 7: Forgetting that the zero function is both even and odd

   For:

   \[ f(x)=0 \]

   both identities hold.

   Therefore, classifying it as even only or odd only is incomplete.

    

   02.3.13 Essential Formula Summary

   Students should memorize the following relationships.

   Essential Parity Formulas

   Even Function

   
   f(−x) = f(x)
   

   Symmetry about the y-axis

   Odd Function

   
   f(−x) = −f(x)
   

   Symmetry about the origin

   Important: For either classification,
   the function’s domain must be symmetric about zero.

   Both even and odd classifications require a symmetric domain.

   02.3.14 Lesson Summary

   Function parity helps students understand algebraic and graphical relationships between positive and negative inputs.

   The most important principles are:

   1. An even function satisfies \(f(-x)=f(x)\) throughout its symmetric domain.

   2. An odd function satisfies \(f(-x)=-f(x)\) throughout its symmetric domain.

   3. Even functions have y-axis symmetry.

   4. Odd functions have origin symmetry.

   5. A function that satisfies neither condition is classified as neither even nor odd.

   6. A function’s domain must be symmetric about zero for even or odd parity.

   7. The zero function is both even and odd.

   8. Polynomial functions containing only even powers are even; those containing only odd powers are odd, provided the polynomial is not identically zero.

   9. Sine and tangent are odd on their natural domains, while cosine is even.

   10. Algebraic substitution and simplification provide a reliable method for testing parity.

   Final exam reminder: Always check the domain before applying an algebraic parity test. The function’s formula alone may not determine its parity when the domain is restricted.